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You have 12 colors of paint. How many *distinct* ways can [#permalink]

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12 Aug 2003, 05:35

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You have 12 colors of paint. How many *distinct* ways can you paint a regular dodecahedron (12-sided 3-d figure) using all 12 colors, one color to a side. (If you can rotate a painted dodecahedron in a manner so that it looks like another painted one, they are the same and not distinct).

Recall a regular dodecahedron has a pentagon on top, one on the bottom and two row of 5 pentagons each in the middle. (if you play D&D, you know what they look like).

You can leave it in combined factorial and/or combinatorial form if desired.

This seems dauntingly complex but is really easy if you understand the method. (I will give at least 3 solutions).
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Re: Challenge: If u can do it u really know your combinatori [#permalink]

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12 Aug 2003, 06:13

mciatto wrote:

AkamaiBrah wrote:

You have 12 colors of paint. How many *distinct* ways can you paint a regular dodecahedron (12-sided 3-d figure) using all 12 colors, one color to a side. (If you can rotate a painted dodecahedron in a manner so that it looks like another painted one, they are the same and not distinct).

Recall a regular dodecahedron has a pentagon on top, one on the bottom and two row of 5 pentagons each in the middle. (if you play D&D, you know what they look like).

You can leave it in combined factorial and/or combinatorial form if desired.

This seems dauntingly complex but is really easy if you understand the method. (I will give at least 3 solutions).

12 * 9!

If you explain your logic I can tell you where you messed up. _________________

Re: Challenge: If u can do it u really know your combinatori [#permalink]

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12 Aug 2003, 06:18

AkamaiBrah wrote:

mciatto wrote:

AkamaiBrah wrote:

You have 12 colors of paint. How many *distinct* ways can you paint a regular dodecahedron (12-sided 3-d figure) using all 12 colors, one color to a side. (If you can rotate a painted dodecahedron in a manner so that it looks like another painted one, they are the same and not distinct).

Recall a regular dodecahedron has a pentagon on top, one on the bottom and two row of 5 pentagons each in the middle. (if you play D&D, you know what they look like).

You can leave it in combined factorial and/or combinatorial form if desired.

This seems dauntingly complex but is really easy if you understand the method. (I will give at least 3 solutions).

I fixed the top side, leaving 11 possibilities for the bottom. Now there are 10 sides in the "middle". For these, from what I have learned in your last post dealing with the cube there are (n-1)! (makes sense because you hold one of them constant and rotate all others) So (10-1)! = 9!

Re: Challenge: If u can do it u really know your combinatori [#permalink]

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12 Aug 2003, 06:37

nice try, but having two layers of 5 is not the same as having one layer of 10. (good start but this problem is slightly -- just slightly -- more involved than the cube problem). Think logically.
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nope. think: what do I need to do the fill the 1st layer, then what do I need to do to fill the 2nd.

One solution is:

assign color#1 to the top surface. There are now 11 other colors that can go on the bottom. For each of these 11 pairs, we need to fill two layers of 5 faces.

Lets fill the layer of 5 faces all adjacent to the top face. We have 10 colors to choose from so there are 10C5 ways to pick 5 colors. Since only the top and bottom were specified before, we started off with rotational symmetry, so there are 4! ways to distinctly color those 5 faces. Now we have 5 faces left and 5 colors. Since the 5 colors we just painted take away the rotation symmetry of the object, we now have 5! ways to paint the last 5 faces. So the answer is:

11 * 10C5 * 4! * 5! = 7983360

Method 2.

assign color#1 to the top surface. This time, lets color the 1st layer of 5 faces. There are 11C5 ways to pick 5 colors from 11 and 4! ways to paint them distinctly (we can rotate about the single top face). Now there are 6C5 ways to color the next 5 faces and 5! ways to make them distinct. The last face is completely determined by everything that happened before. Hence, there are 11C5 * 4! * 10C5 * 5! = 7983360.

Method 3.

Imagine the dodecahefron completely painted with color #1 being on the top face. Well, looking straight down, you can see that there are 5 ways to rotate the object keeping the color#1 face on top. Since there are 12 faces, there are 12x5 = 60 different positions for each distinct pattern.

Since there are 12 faces, there are 12! total different patterns, but every 60 of them are the same distinct pattern, so the answer is 12!/60 = 7983360.
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